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Proof step 35 from proof size-exgr (previous step)

# s1 + 1 < # s0, # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1), ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ # s0 = 0, s0 = ∅, a ∈ s1, a ∈ s0, ∃ a0. a0 ∈ s0 ∧ ¬ a0 ∈ s1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
# s1 + 1 < # s0, # s1 < # s0, a0 ∈ s0, ∀ s0. # s1 < # s0 → (∃ a. a ∈ s0 ∧ ¬ a ∈ s1) ⊦ # s0 = 0, a ∈ s0, a ∈ s1, a0 ∈ s1, ∃ a0. a0 ∈ s0 ∧ ¬ a0 ∈ s1

Rule argument

Additional information

Used simplifier rules: ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%nls-fst: ⊦ ¬ m < n → ¬ m + m0 < n; from specification nat-basic2